MATH 311–504
Spring 2008
Sample problems for the final exam
Any problem may be altered or replaced by a different one!
Problem 1 (15 pts.) The planes x + 2y + 2z = 1 and 4x + 7y + 4z = 5 intersect in a
line. Find a parametric representation for the line.
To find the intersection set, we need to solve the system
x + 2y + 2z = 1,
4x + 7y + 4z = 5.
Let us convert the system to reduced form using elementary operations:
x + 2y + 2z = 1
x + 2y + 2z = 1
x + 2y + 2z = 1
x − 6z = 3
⇐⇒
⇐⇒
⇐⇒
4x + 7y + 4z = 5
−y − 4z = 1
y + 4z = −1
y + 4z = −1
It follows that the general solution of the system is x = 6t + 3, y = −4t − 1, z = t, where t ∈ R.
Therefore the two planes intersect in the line t(6, −4, 1) + (3, −1, 0).
Problem 2 (20 pts.) Consider a linear operator L : R3 → R3 given by
L(v) = (v · v1 )v2 ,
where v1 = (1, 1, 1), v2 = (1, 2, 2).
(i) Find the matrix of the operator L.
Given v = (x, y, z) ∈ R3 , we have that v·v1 = x+y+z and L(v) = (x+y+z, 2(x+y+z), 2(x+y+z)).
Let A denote the matrix of the linear operator L. The columns of A are vectors L(e1 ), L(e2 ), L(e3 ),
where e1 = (1, 0, 0), e2 = (0, 1, 0), e3 = (0, 0, 1) is the standard basis for R3 . Therefore


1 1 1
A = 2 2 2.
2 2 2
(ii) Find the dimensions of the image and the null-space of L.
The image Im L of the linear operator L is the subspace of all vectors of the form L(v), where
v ∈ R3 . It is easy to see that Im L is the line spanned by the vector v2 = (1, 2, 2). Hence dim Im L = 1.
The null-space Null L of the operator L is the subspace of all vectors x ∈ R3 such that L(x) = 0.
Clearly, L(x) = 0 if and only if x · v1 = 0. Therefore Null L is the plane x + y + z = 0 orthogonal to
v1 and passing through the origin. Its dimension is 2.
(iii) Find bases for the image and the null-space of L.
Since the image of L is the line spanned by the vector v2 = (1, 2, 2), this vector is a basis for the
image. The null-space of L is the plane given by the equation x + y + z = 0. The general solution of
the equation is x = −t − s, y = t, z = s, where t, s ∈ R. It gives rise to a parametric representation
t(−1, 1, 0) + s(−1, 0, 1) of the plane. Thus the null-space of L is spanned by the vectors (−1, 1, 0) and
(−1, 0, 1). Since the two vectors are linearly independent, they form a basis for Null L.
1

1
1
Problem 3 (20 pts.) Let A = 
0
0

1
0
0
1
1 −1 
.
1 −2
2
1
0
1
(i) Evaluate the determinant of the matrix A.
The determinant can be evaluated using row or column expansions. For example, let us expand
the determinant of A by the first row:
1 1
0
0 1
1 −1 1
1 −1 1 1
1 −1 = 1 −2
2 − 0 −2
2 .
0 1 −2
2 1
0
1
0
0
1
0 1
0
1
Then expand each of the two 3-by-3 determinants by the third row:
1 −1 1
1
1 1 +
det A = −
= (0 + (−3)) − (−2) = −1.
−2
2 1 −2 0 −2 Another way to evaluate det A is to reduce the matrix A to the identity matrix using elementary
row operations (see below). This requires more work but we are going to do it anyway, to find the
inverse of A.
(ii) Find the inverse matrix A−1 .
First we merge the matrix A with

1
1

0
0
the identity matrix into one 4-by-8 matrix:

1
0
0 1 0 0 0
1
1 −1 0 1 0 0 
.
1 −2
2 0 0 1 0
1
0
1 0 0 0 1
Then we apply elementary row operations to this matrix until the left
matrix.
Subtract the first row from the second row:



1 1
0
0 1 0 0 0
1 1
0
0
1
1 1


1 −1 0 1 0 0 
0 0
1 −1 −1

→
 0 1 −2
 0 1 −2
2 0 0 1 0
2
0
0
0 1
0
1 0 0 0 1
0 1
0
1
Interchange the third row with the second row:



1 0 0 0
1
1 1
0
0


0 0
1 −1 −1 1 0 0 
0

→
0
 0 1 −2
2
0 0 1 0
0 0 0 1
0 1
0
1
0
Subtract the second row from the fourth row:



1
1 1
0
0
1 0 0 0

0
 0 1 −2
0
0
1
0
2
→

0
0 0
1 −1 −1 1 0 0 
0 1
0
1
0 0 0 1
0
2
part becomes the identity
0
1
0
0
1
0
0
1 0
0 0
1 −2
2
0
1 −1 −1 1
0 0
1
0
1
0
0
1
0
0
1
0
0
1
0
0
1 0
0
0 0
1
1 −2
2
0
0
1 −1 −1 1
0
2 −1
0 0 −1

0
0
.
0
1

0
0
.
0
1

0
0
.
0
1
Subtract 2 times

1
0

0
0
the third row from the fourth row:


1
1 0
0 0
1
0
0
0
1 −2
2
0 0
1 0
→
0
0 0
0
1 −1 −1 1
0 0 −1 1
0
2 −1
0
Add the fourth row to the third row:

1 1
0
0
1
0
0
 0 1 −2
2
0
0
1

0 0
1
0
1 −1 −1
0 0
0
1
2 −2 −1
1
0
0
1
0
0
1 −2
2
0
0
1
1
0
0
1 −1 −1
2 −2 −1
0
0
1


0
1


0
0
→
0
0
0
1
Subtract 2 times

1
0

0
0
the fourth row from the second row:


0
0 0
1
0 0 1
1
0
1 −2 2 0
0
1 0
→
0
0
1 0 1 −1 −1 1 
0
0 1 2 −2 −1 1
0
Add 2 times the

1
0

0
0
third row to the second row:
0
0
1
0 0 1
1 −2 2 0
0
1
0
1 0 1 −1 −1
0
0 1 2 −2 −1

0
0
.
1
1

1
0
0
0
1
0 0
1 −2 0 −4
4
3 −2 
.
1 −1 −1
1
0
1 0
0
0 1
2 −2 −1
1


1
0 0
1 1
1
0
0
0
0 1
4
3 −2 
1 −2 0 −4
→
0 0
0
1 0
1 −1 −1
1
2 −2 −1
1
0
0 1
0 0
Subtract the second row from the first row:



1 1 0 0
1
1
0
0 0

 0 1 0 0 −2

2
1 0
0

→
0 0 1 0
0
1 −1 −1 1 
2 −2 −1 1
0 0 0 1
0

0
0
.
0
1
0
1
0
0
0
0
1
0
0
0
1
0
0
1
0
0
2
1
0 −2
0
1 −1 −1
2 −2 −1
1
0
3 −2 −1
2
1
0 −2
0
1 −1 −1
2 −2 −1
1

0
0
.
1
1

0
0
.
1
1
Finally the left part of our 4-by-8 matrix is transformed into the identity matrix. Therefore the
current right side is the inverse matrix of A. Thus
A−1

1
1
=
0
0
−1 

1
0
0
3 −2 −1 0

1
1 −1 
2
1 0
 =  −2
.


1 −2
2
1 −1 −1 1 
1
0
1
2 −2 −1 1
As a byproduct, we can evaluate the determinant of A. We have transformed A into the identity
matrix using elementary row operations. These included one row exchange and no row multiplications.
It follows that det A = − det I = −1.


1 1 1
Problem 4 (25 pts.) Let B =  1 1 1 .
1 1 1
(i) Find all eigenvalues of the matrix B.
3
The eigenvalues of B are roots of the characteristic equation det(B − λI) = 0. One obtains that
1−λ
1
1 det(B − λI) = 1
1−λ
1 = (1 − λ)3 − 3(1 − λ) + 2
1
1
1−λ
= (1 − 3λ + 3λ2 − λ3 ) − 3(1 − λ) + 2 = 3λ2 − λ3 = λ2 (3 − λ).
Hence the matrix B has two eigenvalues: 0 and 3.
(ii) Find a basis for R3 consisting of eigenvectors of B?
An eigenvector x = (x, y, z) of B associated with an eigenvalue λ is a nonzero solution of the vector
equation (B − λI)x = 0. First consider the case λ = 0. We obtain that

   
1 1 1
x
0
Bx = 0 ⇐⇒  1 1 1  y  =  0  ⇐⇒ x + y + z = 0.
1 1 1
z
0
The general solution is x = −t − s, y = t, z = s, where t, s ∈ R. Equivalently, x = t(−1, 1, 0) +
s(−1, 0, 1). Hence the eigenspace of B associated with the eigenvalue 0 is two-dimensional. It is
spanned by eigenvectors v1 = (−1, 1, 0) and v2 = (−1, 0, 1).
Now consider the case λ = 3. We obtain that

   
−2
1
1
x
0
(B − 3I)x = 0 ⇐⇒  1 −2
1  y  =  0 
1
1 −2
z
0
   
1 −1
0
x
0
x − y = 0,
⇐⇒  0
1 −1  y  =  0  ⇐⇒
y − z = 0.
0
0
0
0
z

The general solution is x = y = z = t, where t ∈ R. In particular, v3 = (1, 1, 1) is an eigenvector of B
associated with the eigenvalue 3.
The vectors v1 = (−1, 1, 0), v2 = (−1, 0, 1), and v3 = (1, 1, 1) are eigenvectors of the matrix B.
They are linearly independent since the matrix whose rows are these vectors is invertible:
−1 1 0 −1 0 1 = 3 6= 0.
1 1 1
It follows that v1 , v2 , v3 is a basis for R3 .
(iii) Find an orthonormal basis for R3 consisting of eigenvectors of B?
It is easy to check that the vector v3 is orthogonal to v1 and v2 . To transform the basis v1 , v2 , v3
into an orthogonal one, we only need to orthogonalize the pair v1 , v2 . Namely, we replace the vector
v2 by
1
v2 · v1
v1 = (−1, 0, 1) − (−1, 1, 0) = (−1/2, −1/2, 1).
u = v2 −
v1 · v1
2
Now v1 , u, v3 is an orthogonal basis for R3 . Since u is a linear combination of the vectors v1 and v2 ,
it is also an eigenvector of B associated with the eigenvalue 0.
4
v1
u
v3
, w2 =
, and w3 =
form an orthonormal basis for R3 consisting
|v1 |
|u| √
|v
|
p 3
√
of eigenvectors of B. We get that |v1 | = 2, |u| = 3/2, and |v3 | = 3. Thus
Finally, vectors w1 =
1
w1 = √ (−1, 1, 0),
2
1
w2 = √ (−1, −1, 2),
6
1
w3 = √ (1, 1, 1).
3
Problem 5 (20 pts.) Let V be a three-dimensional subspace of R4 spanned by vectors
x1 = (1, 1, 0, 0), x2 = (1, 3, 1, 1), and x3 = (1, 1, −3, −1).
(i) Find an orthogonal basis for V .
To find an orthogonal basis for the subspace V , we apply the Gram-Schmidt orthogonalization
process to the spanning set x1 , x2 , x3 :
v1 = x1 = (1, 1, 0, 0),
v2 = x2 −
v3 = x3 −
x2 · v1
4
v1 = (1, 3, 1, 1) − (1, 1, 0, 0) = (−1, 1, 1, 1),
v1 · v1
2
−4
x3 · v2
2
x3 · v1
(−1, 1, 1, 1) = (−1, 1, −2, 0).
v1 −
v2 = (1, 1, −3, −1) − (1, 1, 0, 0) −
v1 · v1
v2 · v2
2
4
By construction, the vectors v1 , v2 , v3 are orthogonal to each other. It follows that they are linearly
independent. Also, the span of v1 , v2 , v3 is the same as the span of x1 , x2 , x3 , that is, the subspace
V . Thus v1 , v2 , v3 is an orthogonal basis for V .
(ii) Find the distance from the point y = (2, 0, 2, 4) to the subspace V .
The distance from y to V is the distance from y to the closest point y0 in V , which is the orthogonal
projection of y on V . Since v1 , v2 , v3 is an orthogonal basis for V , it follows that
y0 =
y · v2
y · v3
y · v1
v1 +
v2 +
v3
v1 · v1
v2 · v2
v3 · v3
2
4
−6
= (1, 1, 0, 0) + (−1, 1, 1, 1) +
(−1, 1, −2, 0) = (1, 1, 3, 1).
2
4
6
Then y−y0 = (2, 0, 2, 4)−(1, 1, 3, 1) = (1, −1, −1, 3) and the desired distance is |y−y0 | =
√
√
12 = 2 3.
Bonus Problem 6 (15 pts.) Let S be the set of all points in R3 that lie at same distance
from the planes x + 2y + 2z = 1 and 4x + 7y + 4z = 5. Show that S is the union of two planes
and find these planes.
For any point x0 = (x0 , y0 , z0 ) ∈ R3 let d1 (x0 ) denote the distance from x0 to the plane x+2y+2z =
1 and d2 (x0 ) denote the distance from x0 to the plane 4x + 7y + 4z = 5. Then
d1 (x0 ) =
d2 (x0 ) =
|x0 + 2y0 + 2z0 − 1|
1
√
= |x0 + 2y0 + 2z0 − 1|,
2
2
2
3
1 +2 +2
1
|4x0 + 7y0 + 4z0 − 5|
√
= |4x0 + 7y0 + 4z0 − 5|.
2
2
2
9
4 +7 +4
5
The point x0 belongs to the set S if d1 (x0 ) = d2 (x0 ), i.e., if
1
1
|x0 + 2y0 + 2z0 − 1| = |4x0 + 7y0 + 4z0 − 5|.
3
9
This means that
3(x0 + 2y0 + 2z0 − 1) = 4x0 + 7y0 + 4z0 − 5
or
3(x0 + 2y0 + 2z0 − 1) = −(4x0 + 7y0 + 4z0 − 5).
Equivalently, x0 + y0 − 2z0 = 2 or 7x0 + 13y0 + 10z0 = 8.
Thus the set S is the union of the planes x + y − 2z = 2 and 7x + 13y + 10z = 8.
Bonus Problem 7 (20 pts.) Find a quadratic polynomial that is an orthogonal polynomial relative to the inner product
Z 1
hp, qi =
xp(x)q(x) dx.
0
First observe that for any integers m, n ≥ 0,
Z 1
m n
xm+n+1 dx =
hx , x i =
0
1
.
m+n+2
To get the first three orthogonal polynomials, we apply the Gram-Schmidt orthogonalization process
to the polynomials 1, x, and x2 :
p0 (x) = 1,
p1 (x) = x −
hx, 1i
1/3
2
hx, p0 i
p0 (x) = x −
=x−
=x− ,
hp0 , p0 i
h1, 1i
1/2
3
p2 (x) = x2 −
2
hx2 , p1 i
hx2 , 1i hx2 , p1 i hx2 , p0 i
x−
.
p0 (x) −
p1 (x) = x2 −
−
hp0 , p0 i
hp1 , p1 i
h1, 1i
hp1 , p1 i
3
Evaluating inner products in the latter formula, we obtain that
hx2 , 1i
h1, 1i
=
1/4
1
= ,
1/2
2
1 2 1
1 1
1
2
hx2 , p1 i = hx2 , xi − hx2 , 1i = − · = − = ,
3
5 3 4
5 6
30
2
2
1
2 1 2 2 1
1 2
1
2
hp1 , p1 i = hx, xi − 2 · hx, 1i +
h1, 1i = − 2 · · +
· = − = ,
3
3
4
3 3
3
2
4 9
36
hx2 , p1 i
1/30
6
=
= .
hp1 , p1 i
1/36
5
Thus
2
6
1 4
6
3
1 6
x−
= x2 − x − + = x2 − x +
−
2 5
3
5
2 5
5
10
is the desired orthogonal polynomial.
p2 (x) = x2 −
6
Alternative solution: A quadratic polynomial p(x) = x2 + ax + b is an orthogonal polynomial
if hp, qi = 0 for any polynomial q such that deg q < deg p. Actually, it is enough to require that
hp, 1i = hp, xi = 0. Note that
hp, 1i =
Z
hp, xi =
Z
1
(x3 + ax2 + bx) dx =
1 a b
+ + ,
4 3 2
(x4 + ax3 + bx2 ) dx =
1 a b
+ + .
5 4 3
0
1
0
Hence p(x) is an orthogonal polynomial if and only if the coefficients a and b satisfy the following
system:
a/3 + b/2 = −1/4,
a/4 + b/3 = −1/5.
Solving the system, we obtain
a/3 + b/2 = −1/4
2a + 3b = −1.5
2a + 3b = −1.5
⇐⇒
⇐⇒
a/4 + b/3 = −1/5
3a + 4b = −2.4
a + b = −0.9
⇐⇒
b = 0.3
⇐⇒
a + b = −0.9
Thus p(x) = x2 − 1.2x + 0.3 is an orthogonal polynomial.
7
a = −1.2
b = 0.3